Calculus I (AB): The Fundamental Theorem: Funkified!

If I define the function

such that (i.e. they are real numbers), then evaluate .

Difficulty:

Solution:

This is quite a humdinger. You have to treat a, b, c, and f like numbers even though they don't look like numbers. To start, split the integral into two separate integrals--they are complex enough by themselves!

Your only variable (for now) is t. We'll worry about the x later. The integral on the left is as easy as falling off a chair. Since it's a variable to a constant exponent (remember, c is constant), add one to it and divide by the new power. More complex is the right integral. You have to use u-substitution with u = 3ft. When you derive u you get du = 3f dt. (Only the t disappears whereas the coefficient 3f remains...it's just like taking the derivative of 12x and getting 12.) Danger! When you do u-substitution, you have to change the boundaries of the definite integral. The old boundaries of a and 2x will have to plugged separately into your u equation for t. Thus, the a boundary becomes u = 3f(a) and the other becomes u = (3)(f)(2x)=6fx:

That fraction in front of the second term is a result of u-substitution. In order to replace dt with du, you have to divide the equation by 3f. Also, note that the boundaries only changed on the integral for which we u-substituted. That's why we split them up in the first step.

Now we have to apply the Fundamental Theorem by plugging the boundaries into t and subtracting as appropriate:

That's f(x). To find the function values required by the problem, plug in for x. Two of the terms will vanish when you find the derivative--the second and fourth terms are all just constants, and the constant of a derivative is 0:

You can't do any other simplification; we don't know the values of the variables.

Thanks to Mr. Sweeny's class in Berlin High School, in Berlin, NH, for their participation in this problem!

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Find the antiderivative of .

Difficulty:

Solution:

The title kind of ruins the problem--you know it's integration by parts. Here's a good tip: when picking u and dv values, choose a u whose derivative is eventually 0, if it's at all possible. Why do this? Because you can then apply the shortcut--the tabular method (i.e. you use a table). In this problem, , so du = 2x. That leaves , meaning . The first column is u followed by subsequent derivatives until you reach 0. The second column is dv and subsequent integrals. Stop integrating dv in the row when u turns into 0. The last column is an alternating +1 and -1. The table always starts with a +1. Also, add an extra member of this column below the table. You get the following:

Multiply along the arrows, which begin in the top- and leftmost cell and proceed diagonally down and to the right. Each arrow indicates a term of the answer. Again, just multiply the cells of the chart together. You will get:

Did you forget the "+C"? It's still an indefinite integral we're doing (no integration boundaries), so you need that C.